1
$\begingroup$

Let $V$ be an affine algebraic variety defined over $\mathbb R$. We assume that $V\subset \mathbb A_n$(affine $n$ space). Suppose that for any algebraic curve $C$ in $V$ defined over $\mathbb R$, the real points $C(\mathbb R)$ is contained in a finite union of proper affine $\mathbb R$-subspaces of $\mathbb A_n$. Is it true that $V(\mathbb R)$ is contained in a finite union of affine $\mathbb R $-subspaces of $\mathbb A_n$.

$\endgroup$
4
  • 2
    $\begingroup$ If I'm understanding the question correctly, the answer is no: what if $V$ is a real plane cubic? $\endgroup$
    – user5117
    Oct 27, 2013 at 9:32
  • $\begingroup$ @Artie: Maybe you are right, but that's not what I want to know. The question is modified. $\endgroup$
    – ronggang
    Oct 27, 2013 at 10:21
  • $\begingroup$ Dear ronggang, ok. In that case, it seems like another title might be more suitable. $\endgroup$
    – user5117
    Oct 27, 2013 at 10:59
  • $\begingroup$ @Artie: I agree. It seems rational relates to open subsets of affine space in algebraic geometry. $\endgroup$
    – ronggang
    Oct 27, 2013 at 11:23

1 Answer 1

3
$\begingroup$

Yes. We can assume without loss of generality that the real points of $V$ are Zariski dense. Choose some linear projection $\rho: V \to \mathbb R^{\dim V}$ whose image includes some open ball, e.g. a generic linear projection. Then if $L$ is a linear subspace of codimension $1$ in $\mathbb R^n$, $\rho(V \cap L )$ is a variety of dimension at most $\dim V -1$. So we have a $n$-dimensional family of subvarieties of codimension $1$ of this open ball.

Choose a finite set of points of the open ball which is on none of these subvarieties.

Choose a generic curve of high degree passing through the finite set of points. Such a curve will have infinitely many real points, and be irreducible. Or just choose a parametrized curve passing through those points. Since it does not lie in $\rho(V \cap L)$, for any $L$, and is irreducible it intersects $\rho(V \cap L)$ at only finitely many points, so it is contained in no finite union.

Then the inverse image of this curve in $V$ will be contained in the union of no finite set of linear subspaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.